# Prove that the set {X: XA = AX} is a subspace of M 2,2

## Homework Statement

A is a 2 x 2 matrix. Prove that the set W = {X: XA = AX} is a subspace of M2,2

## The Attempt at a Solution

Non-emptiness:

Code:
W must be non-empty because the identity matrix I is an element of W.
IA = AI
A = A

Code:
Let X, Y be elements of W such that XA = AX, YA = AY. Add the equations together.
(X+Y)A = A(X+Y)
XA+YA = AX+AY

Hence, X+Y is an element of W.
I have no idea what to do with scalar multiplication. This is my current attempt:

Code:
Let X be an element of W, and c be a scalar.
cXA = AcX
(cX)A = A(cX)
I also tried this:

Code:
Let X be an element of W, and c be a scalar.
XA = AX
c(XA) = c(AX)
I don't feel like either of these attempts proves anything.

I don't understand how to prove that cX is actually in W. This doesn't make any sense to me. How do I prove that?

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MathematicalPhysicist
Gold Member
It's too trivial.
You have X in W, thus XA=AX, thus cXA=A(cX) meaning cX in W.
Or in other word you've proved it already.

It's too trivial.
You have X in W, thus XA=AX, thus cXA=A(cX) meaning cX in W.
Or in other word you've proved it already.
But how do I know that there isn't a scalar that would change X in such a way that (cX)A ≠ A(cX)?

Edit: I guess I get what you're saying now, but it still confuses me how I can just say that they're equal with no real evidence.

MathematicalPhysicist
Gold Member
They are equal cause you have XA=AX, and then you multiply this equation by c.
I said it's trivial.

MathematicalPhysicist
Gold Member
Remember that your scalar is some number, real/complex or something else.
It's not a vector (in which case multiplication of a matrix with a vector will not yield you a matrix of the same size but a vector, and thus it wouldn't be a vector space.

Yeah, I understand it now. This is the first class I've had to write my own proofs in, so I'm still getting used to it. Thanks again for the help.